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Communist · May 19, 2011, 12:47 AM

Can someone please tell me whether these are correct / how to fix them if they aren't?

(a) $(a-b)^8$

$= a^8 - 8a^7b - 28a^6b^2 - 56a^5b^3 - 70a^4b^4 - 56a^3b^5 - 28a^2b^6 - 8ab^7 + b^8$

(b) $(3-2x)^8$

$= 3^8 - 8*3^7*2x - 28*3^6*2x^2 - 56*3^5*2x^3 - 70*3^4*2x^4 - 56*3^3*2x^5 - 28*3^2*2x^6 - 8*3*2x^7 + 2x^8$
$= 6561 - 34992x - 40824x^2 - 27216x^3 - 11340x^4 - 3024x^5 - 504x^6 - 48x^7 + 2x^8$

jcaron2 · May 19, 2011, 06:54 AM

Close, but you have a couple of mistakes. In the first expression, you got all of the coefficients right, but many of the signs are wrong. If it was $(a+b)^8$ , all the coefficients would be positive, of course. Since there's a negative sign in front of the b, however, all the terms with odd powers of b will be negative. Hence it would be $a^8-8a^7b+28a^6b^2-56a^5b^3+70a^4b^4-...$ .

In the second case, you have the same issue with the signs, plus you also have to make sure you include the 2 when raising 2x to various powers. Hence, it should be $(2x)^3=8x^3$ , for example.

Communist · May 20, 2011, 05:40 PM

Originally Posted by jcaron2

Close, but you have a couple of mistakes. In the first expression, you got all of the coefficients right, but many of the signs are wrong. If it was $(a+b)^8$ , all the coefficients would be positive, of course. Since there's a negative sign in front of the b, however, all the terms with odd powers of b will be negative. Hence it would be $a^8-8a^7b+28a^6b^2-56a^5b^3+70a^4b^4-...$ .

In the second case, you have the same issue with the signs, plus you also have to make sure you include the 2 when raising 2x to various powers. Hence, it should be $(2x)^3=8x^3$ , for example.

Thanks! That was very helpful.

jcaron2 · May 20, 2011, 06:16 PM

Originally Posted by Communist

Thanks! That was very helpful.

You're welcome!